If the sum of the coefficients of the first three terms in the expansion of $\left(x-\dfrac{a}{x^2}\right)^{12}, x \neq 0$ is 559. Find the value of '$a$' if '$a$' belongs to positive integers

The expansion of $\left(x - \dfrac{a}{x^2}\right)^{12}$ is given by the binomial theorem as $\sum_{r=0}^{12} {^{12}C_{r}} (x)^{12-r} \left(-\dfrac{a}{x^2}\right)^{r}$.

The first three terms correspond to $r=0, 1, 2$.

For $r=0$: Term is ${^{12}C_{0}} (x)^{12} (-a)^{0} (x)^{-0} = 1 \cdot x^{12}$. The coefficient is $1$.

For $r=1$: Term is ${^{12}C_{1}} (x)^{11} (-a)^{1} (x)^{-2} = 12 \cdot (-a) \cdot x^{9} = -12a x^{9}$. The coefficient is $-12a$.

For $r=2$: Term is ${^{12}C_{2}} (x)^{10} (-a)^{2} (x)^{-4} = 66 \cdot a^{2} \cdot x^{6} = 66a^{2} x^{6}$. The coefficient is $66a^{2}$.

The sum of these coefficients is given as $559$.

$1 - 12a + 66a^{2} = 559$

$66a^{2} - 12a - 558 = 0$

Dividing by $6$: $11a^{2} - 2a - 93 = 0$.

Solving the quadratic equation using the quadratic formula $a = \dfrac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$:

$a = \dfrac{2 \pm \sqrt{4 - 4(11)(-93)}}{22} = \dfrac{2 \pm \sqrt{4 + 4092}}{22} = \dfrac{2 \pm \sqrt{4096}}{22} = \dfrac{2 \pm 64}{22}$.

Since $a$ is a positive integer, $a = \dfrac{2 + 64}{22} = \dfrac{66}{22} = 3$.

Answer: 3